I am sure that this question has been asked before, but what is the solution of $y_i = \frac{e^{x_i}}{\Sigma_{j=1}^{n} e^{x_j}} \forall i $? i.e. how does one make $x_i$ the subject?
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Note that the exponential is no real complication here---by setting $z_i := \exp x_i$ we may as well ask about $y_i = \frac{z_i}{\sum_i z_i}$. – Travis Willse Jun 03 '17 at 19:46
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Also, note that the vector $(z_i)$ can only be recovered up to a nonzero constant, as $(z_i)$ and $(\lambda z_i)$, $\lambda \neq 0$, will yield the same $(y_i)$. – Travis Willse Jun 03 '17 at 19:47
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Let $S = \sum_j e^{x_j}$ then we have $Sy_i = \exp(x_i) \forall i$.
Therefore $x_i = \log(y_i) + \log(S)$.
It is quite easy to see that we cannot determine $S$ for $n\geq 2$, because we basically have an injective map, or in other words, the $x_i$ we have $n$ degrees of freedom, but the $y_i$ only have $n-1$ degrees of freedom as they will always satisfy $ \sum_i y_i = \sum_i \frac{e^{x_i}}{S} = \frac{S}{S} = 1$
flawr
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Sorry all - that was a clear brainfart. Yes, I agree with Travis' comments. To answer my own question, I would fix $z_0$ (where $z_i = e^{x_i}$) and then solve for all the other $z_i$s using the known $y_i$s.
tangerine
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